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Determining Reaction Orders and Rate Constants
Determining Reaction Orders with Empirical Data Consider a reaction 3A + B + 2C --> D + 2E with a rate law Rate = kAmBn Find the rate law, then find the intial rate when A = B = 1.00 M Even though this is not a very well designed experiment, note that the rate law is still solvable with the given data as long as there are is one more experiment than number of reactants, with at least one (in a well-designed experiment, only one) concentration changing. These problems simply involve taking the ratio of the rates and concentrations of two experiments, and this procedure is done multiple times to figure out the reaction orders. Rate 2 = Am2Bn2 Rate 1 = Am1Bn1 1.00 x 10-3 = (0.2)m2(0.5)n2 1.25 x 10-4 = (0.1)m1(0.25)n1 8 = (2)m(2)n log28 = log2((2)m(2)n) 3 = m + n m = 3 - n Rate 3 = Am3Bn3 Rate 1 = Am1Bn1 4.00 x 10-3 = (0.4)m3(0.5)n3 1.25 x 10-4 = (0.1)m1(0.25)n1 32 = (4)m(2)n log232 = log2((4)m(0.5)n(2)p) log2(2)5 = log2(2)2m + log2(2)n 5 = 2m + n Now we have enough information to find the reaction orders (using equations in bold) 5 = 2(3 - n) + n 5 = -n + 6 -1 = -n n = 1 5 = 2m + 1 4 = 2m m = 2 Therefore, Rate = kA2B Finding the Numerical Rate Constant Calculations To find the k value, plug in the values for one of the experiments. Units can be calculated separately, but it is less work and less confusing if they are calculated along with the numerical value. The units can be memorized as well with a formula. 1.25 x 10-4 mol/L∙s = k(0.1 mol/L)2(0.25 mol/L) k = 0.05 (mol/L∙s)/(mol/L)3 k = 0.05 L2/mol2∙s Final Rate Equation: Rate = 0.05A2B Finally, the problem asks to find the rate when A=B=1 M Rate = 0.05(1)2(1) Rate = 0.05 mol/L∙s Formula: Units of k = (L/mol)order - 1 ∙ (units of t)-1 Integrated Rate Laws Rate laws described before were simply initial rate laws and depended on only concentration and not time. However, it is useful to know the concentration of each reactant or product after a certain amount of time. By using intengrated rate laws, we can answer such questions as how much reactant will be used up at a certain time. By redefining a rate as change in concentration over change in time and setting this equal to the corresponding rate law, then integrating over time, an integrated rate law can be determined for all order reactions. The formulas for orders 0, 1, and 2 are below. Note: When using these formulas, be careful about the signs of the right-side terms (mainly k) and the sign of the slope. '2nd order:' Initial Rate Law: Rate = kA2 1/At - 1/A0 = kt '1st order:' Initial Rate Law: Rate = kA lnA0 - lnAt = kt 'Oth order:' Initial Rate Law: Rate = kA0 = k At - A0 = -kt Determining Reaction Order with Graphical Data Say perhaps that you are trying to determine the rate law for a reaction and don't have the initial rates. You could use a graphical method to determine the reaction order, instead, and this means that you don't have to run three or four different experiments just to get the rate law. By analyzing the concentration over time, only one experiment is required. This method relies on using the integrated rate laws, even though the reaction order is not known. ]Given that the reaction order is 0, 1, or 2, graphing A, lnA, and 1/A will result in one of the three graphs being linear. The specific graph that corresponds to a specific reaction order is the same as those for the integrated rate laws, as these methods are derived from those integrated rate laws. They are restated in the following table. Practice Problems Problem 1 Given the reaction A + 2B + 4C --> D + 2E (a) Find the rate law with k value and units (b) determine the initial rate with A = 1.000 M, B = 0.250 M, and C = 2.000 M © determine the integrated rate law (d) determine how much of compound E will be produced after 1 minute